|q⟩ Bad Qubits

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Multiple choice
In the Deutsch-Jozsa algorithm cast as an HSP over G=Z2nG = \mathbb{Z}_2^n, the final HnH^{\otimes n} gives the all-zeros outcome amplitude α=(1/2n)x(1)f(x)\alpha = (1/2^n) \sum_x (-1)^{f(x)}. What does measuring 0n0^n tell you?